Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

f(x) be a continuous functions on the interval [a, b]

and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a).

Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0.

In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = f'(c) = 0).

Rolle's Theorem

If f(x) is

1) a continuous function on the interval [a, b]

2) differentiable on the open interval (a, b)

3) and f(a) = f(b),

then there is at least one value c of x in the interval (a, b) such that

f '(c) = 0

1) a continuous function on the interval [a, b]

2) differentiable on the open interval (a, b)

3) and f(a) = f(b),

then there is at least one value c of x in the interval (a, b) such that

f '(c) = 0

f '(x) = - 2 x + 6

f '(c) = - 2 c + 6 = 0

Solve the above equation to obtain

c = 3

Therefore at x = 3 there is a tangent to the graph of f that has a slope equal to zero (horizontal line) as shown in figure 1 below.

f '(x) = cos(x)

f '(c) = cos(c) = 0

The above equation has two solutions on the interval [0 , 2?]

c

Therefore both at x = ?/2 and x = 3 ?/2 there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below.

a) f(x) = cos(x) , for x in [0 , 2?]

b) g(x) = |x - 2| , for x in [0 , 4]

c) h(x) = 1 / x

d) k(x) = |sin(x)| , for x in [0 , 2?]

f(0) = 1 and f(2?) = 1 therefore f(0) = f(2?)

f is continuous on [0 , 2?]

Function f is differentiable in (0 , 2?)

Function f satisfies all conditions of Rolle's theorem

b)

function g has a V-shaped graph with vertex at x = 2 and is therefore not differentiable at x = 2.

Function g does not satisfy all conditions of Rolle's theorem

c)

Function h is undefined at x = 0.

Function h does not satisfy all conditions of Rolle's theorem.

d)

k(x) = |sin(x)| , for x in [0 , 2?]

The graph of function k is shown below and it shows that function k is not differentiable at x = ?.

Function k does not satisfy all conditions of Rolle's theorem.

f '(x) = 2 x - 4

f '(c) = 2 c - 4 = 0

Solve for c to obtain

c = 2.

The graph below shows f , f ' and the tangent at x = c = 2 is horizontal; and f '(2) = 0

(-?/2 , 3?/2). Also g(- ?/2) = g(3?/2) = 0 and therefore function g satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0.

g '(x) = - sin(x)

g '(c) = - sin(c) = 0

Solve for c to obtain

c = n? , n = 0,± 1 , ± 2 , ...

Solutions on the interval [- ?/2 , 3?/2] are

c

The graph below shows g , g ' and the tangents at x = c

Continuous Functions in Calculus

Non Differentiable Functions

calculus